Let's solve the quadratic equation $4x^2 - 4x + 1 = 0$ using the steps provided.

Step 1: Determine the values of $a$, $b$, and $c$.

The standard form of a quadratic equation is $ax^2 + bx + c = 0$. Comparing this to the given equation $4x^2 - 4x + 1 = 0$:

• $a = 4$
• $b = -4$
• $c = 1$

Step 2: Substitute the values of $a$, $b$, and $c$ in the quadratic formula.

The quadratic formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substitute $a = 4$, $b = -4$, and $c = 1$ into the formula: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(1)}}{2(4)}$ $x = \frac{4 \pm \sqrt{16 - 16}}{8}$ $x = \frac{4 \pm \sqrt{0}}{8}$

Step 3: Solve to find the solutions.

Since $\sqrt{0} = 0$, the equation simplifies to: $x = \frac{4 \pm 0}{8}$ This gives only one solution: $x = \frac{4}{8} = \frac{1}{2}$

Thus, the solution to the quadratic equation is $x = \frac{1}{2}$.

Step 4: Application of the Steps

After applying the quadratic formula, we found that the only solution is $x = \frac{1}{2}$.

Step 5: Check by substituting the solution in the original equation.

Now, let's substitute $x = \frac{1}{2}$ into the original equation $4x^2 - 4x + 1 = 0$:

$4\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right) + 1 = 0$ $4 \times \frac{1}{4} - 4 \times \frac{1}{2} + 1 = 0$ $1 - 2 + 1 = 0$ $0 = 0$ The solution checks out, as the left side equals the right side.

Conclusion:

The solution to the equation $4x^2 - 4x + 1 = 0$ is $x = \frac{1}{2}$.

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Would you like more details or have any questions?

Here are 5 related questions:

1. What happens if the discriminant $b^2 - 4ac$ is negative in the quadratic formula?
2. How would the quadratic equation change if $b = 0$?
3. Can a quadratic equation have no real solutions? How would that happen?
4. What is the geometric interpretation of a quadratic equation's solution on a graph?
5. Can you solve a quadratic equation by factoring instead of using the quadratic formula?

Tip: The discriminant $b^2 - 4ac$ determines the number and type of solutions: if it's positive, you get two real solutions; if it's zero, one real solution; and if it's negative, no real solutions (only complex).